## Change of basis (Matrix and operator)

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Ddanndt
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### Change of basis (Matrix and operator)

Hi everybody. I've been studying vector spaces for a few weeks and we're now doing change of basis but I keep getting the matrix and operators wrong. So can someone please explain the new basis/ old basis thing clearly please?
God does not care about our mathematical difficulties — He integrates empirically.
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gorcee
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### Re: Change of basis (Matrix and operator)

Let's take a 2-D example. Say you have an x-y coordinate frame. If you have a vector $(v_1, v_2)$ in this coordinate frame, then it can be written as a linear combination of the standard basis vectors, $e_1 = (1,0)$ and $e_2 = (0,1)$.

Transforming something to a new basis just means that you're using a different set of basis vectors to write the same thing. That is, you want your vector to be the same, but you're referring to it using a different coordinate system.

The key idea here is that a vector is always able to be written as a linear combination of basis vectors. If, instead of the standard basis, you had a basis $\hat{e}_1 = (1,2)$, $\hat{e}_2 = (3,-1)$, then you could write $(v_1,v_2) = c_1\hat{e}_1 + c_2\hat{e_2}$. If you write out the system fully, you'll get a linear system (hint, the first equation is $v_1 = c_1 \hat{e}_{11} + c_2 \hat{e}_{21}$) and you can compute $c_1$ and $c_2$.

PM 2Ring
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### Re: Change of basis (Matrix and operator)

Double \$ signs work better:

gorcee wrote:Let's take a 2-D example. Say you have an x-y coordinate frame. If you have a vector $$(v_1, v_2)$$ in this coordinate frame, then it can be written as a linear combination of the standard basis vectors, $$e_1 = (1,0)$$ and $$e_2 = (0,1)$$.

Transforming something to a new basis just means that you're using a different set of basis vectors to write the same thing. That is, you want your vector to be the same, but you're referring to it using a different coordinate system.

The key idea here is that a vector is always able to be written as a linear combination of basis vectors. If, instead of the standard basis, you had a basis $$\hat{e}_1 = (1,2)$$, $$\hat{e}_2 = (3,-1)$$, then you could write $$(v_1,v_2) = c_1\hat{e}_1 + c_2\hat{e_2}$$. If you write out the system fully, you'll get a linear system (hint, the first equation is $$v_1 = c_1 \hat{e}_{11} + c_2 \hat{e}_{21}$$) and you can compute $$c_1$$ and $$c_2$$.

squareroot1
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### Re: Change of basis (Matrix and operator)

Are you getting the matrix entirely wrong, or just "backwards" (where you would swap the old and new basis or invert the matrix to get the correct value)? I remember always writing it the wrong-way-round and getting hung up cause of that.

To work with the numbers already given, label A the change of basis matrix from [imath]e_i[/imath] to [imath]f_i[/imath]. We know that [imath]f_1=e_1+2e_2[/imath] and [imath]f_2=3e_1-e_2[/imath]. This means that A has to send (1,2) onto (1,0) and (3,-1) onto (0,1). Find an expression for A which does that (Hint: consider combining the input and output vectors into matrixes,) and you will have your change of basis matrix.

gorcee
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### Re: Change of basis (Matrix and operator)

Gah thanks PM... too much time spent on MathOverflow, not enough previewing before posting.

aleph_one
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### Re: Change of basis (Matrix and operator)

Is there a common notation for matrices and vectors in different bases that make clear active and passive coordinate transformations?

Yakk
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### Re: Change of basis (Matrix and operator)

$$(a,b,c,d)_B$$ for the vector in basis B?
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.